Conceptual doubt about quadratic inequalities?

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    i am learning how to solve quadratic inequalities graphically but it is the same steps every single time - i dont want to learn algorithms i want to learn maths so my doubt is
    1) Why should the inequality be written with 0 alone on one side like ax^2+bx+c<0

    2) What is the reason that makes the x axis taken to be that 0
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    1) Solving for ax^2+bx+c=0 will give you the so-called critical values of x, in other words the values of x at which the expression transitions between being <0 and >0.

    2) y=0 is just another way of identifying the x axis.
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    (Original post by tryingtomaths)
    i am learning how to solve quadratic inequalities graphically but it is the same steps every single time - i dont want to learn algorithms i want to learn maths so my doubt is
    1) Why should the inequality be written with 0 alone on one side like ax^2+bx+c<0

    2) What is the reason that makes the x axis taken to be that 0
    If you've been taught this as a set of steps with no explanation then you're being taught badly.


    Let's say you're solving the inequality x^2 + 3x - 4 &lt; 0

    Then consider the function below

    y  = x^2 + 3x - 4


    Since y is equal to x^2 + 3x - 4 \ \ , if y &lt; 0 then this gives us the original inequality

    x^2 + 3x - 4 &lt; 0


    So solving x^2 + 3x - 4 &lt; 0 is equivalent to finding where y is &lt; 0 in the graph of y  = x^2 + 3x - 4.

    To find this, you first need to find where x^2 + 3x - 4 = 0 to give you the so called critical values where y switches from positive to negative (or vice-versa).


    The reason why you start by rearranging into the form ax^2+bx+c &lt; 0 is because you'll at some point need to solve the corresponding quadratic equation as shown above. And to do this you have to put all the terms on one side. You will have learnt this at GCSE.
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    (Original post by notnek)
    If you've been taught this as a set of steps with no explanation then you're being taught badly.


    Let's say you're solving the inequality x^2 + 3x - 4 &lt; 0

    Then consider the function below

    y  = x^2 + 3x - 4


    Since y is equal to x^2 + 3x - 4 \ \ , if y &lt; 0 then this gives us the original inequality

    x^2 + 3x - 4 &lt; 0


    So solving x^2 + 3x - 4 &lt; 0 is equivalent to finding where y is &lt; 0 in the graph of y  = x^2 + 3x - 4.

    To find this, you first need to find where x^2 + 3x - 4 = 0 to give you the so called critical values where y switches from positive to negative (or vice-versa).


    The reason why you start by rearranging into the form ax^2+bx+c &lt; 0 is because you'll at some point need to solve the corresponding quadratic equation as shown above. And to do this you have to put all the terms on one side. You will have learnt this at GCSE.
    Thank you so much for your beautiful explanation; yes so far i can follow up, but i still don't understand why we linked the "y" from the function to replace the original question; yes i realize they are equal but so did the equation when we made it equal to 0; yet we did'nt say that 0< 0; what made us choose to say that y < 0? sorry if my question is stupid
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    (Original post by tryingtomaths)
    Thank you so much for your beautiful explanation; yes so far i can follow up, but i still don't understand why we linked the "y" from the function to replace the original question; yes i realize they are equal but so did the equation when we made it equal to 0; yet we did'nt say that 0< 0; what made us choose to say that y < 0? sorry if my question is stupid
    Firstly, solving x^2+3x-4=0 is part of the method for solving the inequality.

    I am not actually saying that solving x^2+3x-4&lt;0 is the same as solving x^2+3x-4=0.

    You're right : if both of these are true : x^2+3x-4&lt;0 and x^2+3x-4=0 then you could derive 0&lt;0, which doesn't make sense.


    If you're unsure of the above, try to ignore it for now and focus on the following:

    What I am saying is that solving x^2+3x-4&lt;0

    is equivalent to solving this pair

    y=x^2+3x-4 and  y&lt;0.


    The reason these are equivalent is because writing y=x^2+3x-4 means that y is the same as x^2+3x-4.

    So if  y&lt;0 and y is the same as x^2+3x-4 then it must be true that

    x^2+3x-4&lt;0


    If you're still unsure, please let me know the exact line of my post where you first were confused.
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    (Original post by notnek)
    Firstly, solving x^2+3x-4=0 is part of the method for solving the inequality.

    I am not actually saying that solving x^2+3x-4&lt;0 is the same as solving x^2+3x-4=0.

    You're right : if both of these are true : x^2+3x-4&lt;0 and x^2+3x-4=0 then you could derive 0&lt;0, which doesn't make sense.


    If you're unsure of the above, try to ignore it for now and focus on the following:

    What I am saying is that solving x^2+3x-4&lt;0

    is equivalent to solving this pair

    y=x^2+3x-4 and  y&lt;0.


    The reason these are equivalent is because writing y=x^2+3x-4 means that y is the same as x^2+3x-4.

    So if  y&lt;0 and y is the same as x^2+3x-4 then it must be true that

    x^2+3x-4&lt;0


    If you're still unsure, please let me know the exact line of my post where you first were confused.
    what i still don't understand and can't wrap my head around is why and how we derived that when y=x^2+3x-4; y moves up and we find the inequality with respect to it; however when we put some other number the same rule doesn't apply. ; i even read the purplemaths article on quadratic inequalities and understood it well but still stuck at that same problem.Thank you so much for your patience with me.
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    (Original post by tryingtomaths)
    what i still don't understand and can't wrap my head around is why and how we derived that when y=x^2+3x-4; y moves up and we find the inequality with respect to it; however when we put some other number the same rule doesn't apply. ; i even read the purplemaths article on quadratic inequalities and understood it well but still stuck at that same problem.Thank you so much for your patience with me.
    I'm not sure what you mean by "y moves up". Can you explain this with an example?

    Was there any part of my last post that didn't make sense?
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    (Original post by notnek)
    I'm not sure what you mean by "y moves up". Can you explain this with an example?

    Was there any part of my last post that didn't make sense?
    I mean that y replaces that main equation and becomes y<0.
    And yes
    if both of these are true : x^2+3x-4&lt;0 and x^2+3x-4=0 then you could derive 0&lt;0, which doesn't make sense
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    (Original post by tryingtomaths)
    I mean that y replaces that main equation and becomes y<0.
    And yes
    Ignore that sentence you quoted - it's not important. The part below that sentence is where I've tried to explain your issue. Can you tell me which line make sense?

    It would be useful if you start with x^2 + 3x - 4 < 0 and go through the method of solving it (post all the working in full) and then stop at the exact point where you beome unsure of what's going on.

    I recommend you look for videos on this topic. Hopefully someone else explaining it visually will help.
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    (Original post by notnek)
    Ignore that sentence you quoted - it's not important. The part below that sentence is where I've tried to explain your issue. Can you tell me which line make sense?

    It would be useful if you start with x^2 + 3x - 4 < 0 and go through the method of solving it (post all the working in full) and then stop at the exact point where you beome unsure of what's going on.

    I recommend you look for videos on this topic. Hopefully someone else explaining it visually will help.
    Very well;
    • x2+3x-4<0
    • (x+4)(x-1)<0
    • y=(x+4)(x-1)
    • 0=(x+4)(x-1)
    • x=-4 l x=1
    • since y<0
    • -4<x<1
    i know that we are basically looking at when the parabola y=x2+3x-4 is below the line y=0 and that we will then equate the equation to 0 to find out the points of intersection with x axis. My only doubt is on what basis did we determine that the 0 is connected to the line y=0? you say it's because the graph function would be y= x2+3x-4 but then again 0 was also equal to the equation and we did'nt say 0<0. i looked up at many youtube videos which included hegartymaths; ukmathsteacher and examsolutions but none explained why. I am so sorry for taking up your time
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    (Original post by tryingtomaths)
    Very well;
    • x2+3x-4<0
    • (x+4)(x-1)<0
    • y=(x+4)(x-1)
    • 0=(x+4)(x-1)
    • x=-4 l x=1
    since y<0-4<x<1; i know that we are basically looking at when the parabola y=x2+3x-4 is below the line y=0 and that we will then equate the equation to 0 to find out the points of intersection with x axis. My only doubt is on what basis did we determine that the 0 is connected to the line y=0? you say it's because the graph function would be y= x2+3x-4 but then again 0 was also equal to the equation. i looked up at many youtube videos which included hegartymaths; ukmathsteacher and examsolutions but none explained why. I am so sorry for taking up your time
    Thank you for that. Since you've watched all the videos without understanding, an explanation on an internet forum may not be enough. It would benefit you to have a face to face explanation of this where the you can go back and forth with the teacher until you understand.

    I'm going to repost my explanation of what I think is your question (let me know if it isn't). Please tell me if there's anything that doesn't make sense.

    What I am saying is that solving x^2+3x-4&lt;0

    is equivalent to solving this pair

    y=x^2+3x-4 and  y&lt;0.


    The reason these are equivalent is because writing y=x^2+3x-4 means that y is the same as x^2+3x-4.

    So if  y&lt;0 and y is the same as x^2+3x-4 then it must be true that

    x^2+3x-4&lt;0


    I would be grateful if someone else reading this thread can attempt to explain.
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    (Original post by notnek)
    Thank you for that. Since you've watched all the videos without understanding, an explanation on an internet forum may not be enough. It would benefit you to have a face to face explanation of this where the you can go back and forth with the teacher until you understand.

    I'm going to repost my explanation of what I think is your question (let me know if it isn't). Please tell me if there's anything that doesn't make sense.

    What I am saying is that solving x^2+3x-4&lt;0

    is equivalent to solving this pair

    y=x^2+3x-4 and  y&lt;0.


    The reason these are equivalent is because writing y=x^2+3x-4 means that y is the same as x^2+3x-4.

    So if  y&lt;0 and y is the same as x^2+3x-4 then it must be true that

    x^2+3x-4&lt;0


    I would be grateful if someone else reading this thread can attempt to explain.
    yes this is exactly my question and the explanation makes senses; and because it make's sense i am asking why can't we say 0<0 if we have x^2+3x-4=0 in our steps? i am really not stupid i just question alot
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    (Original post by tryingtomaths)
    yes this is exactly my question and the explanation makes senses; and because it make's sense i am asking why can't we say 0<0 if we have x^2+3x-4=0 in our steps? i am really not stupid i just question alot
    Solving x^2+3x-4=0 is useful so that we can draw y = x^2 + 3x - 4 and then use the graph to solve x^2+3x-4<0

    We're not acutually saying that x^2+3x-4=0. It's just part of the method.

    You're not stupid. Actually you're the kind of student I like to teach - someone who asks questions.
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    (Original post by notnek)
    Solving x^2+3x-4=0 is useful so that we can draw y = x^2 + 3x - 4 and then use the graph to solve x^2+3x-4<0

    We're not acutually saying that x^2+3x-4=0. It's just part of the method.

    You're not stupid. Actually you're the kind of student I like to teach - someone who asks questions.
    I think i got it. So basically the function doesn't =0; we equate it to 0 just to find the points of intersection at y=0, while the whole curve will be equal to y. and by saying 0<0 we are only considering those points at the x axis rather than the whole parabola. Thank you i got it now; it's been 2 days i am trying to understand it. and thank you so much for your kind comment. you made my day
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    (Original post by tryingtomaths)
    I think i got it. So basically the function doesn't =0; we equate it to 0 just to find the points of intersection at y=0, while the whole curve will be equal to y. and by saying 0<0 we are only considering those points at the x axis rather than the whole parabola. Thank you i got it now; it's been 2 days i am trying to understand it. and thank you so much for your kind comment. you made my day
    I'm not completely sure what you mean when you talk about "0<0" but if it all makes sense to you then that's what matters.

    Make sure you continue to ask questions on here if there's something like this that wasn't clear to you. There are plenty of members here who will take a lot of time to explain a concept and won't stop until you're happy
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    (Original post by notnek)
    I'm not completely sure what you mean when you talk about "0<0" but if it all makes sense to you then that's what matters.

    Make sure you continue to ask questions on here if there's something like this that wasn't clear to you. There are plenty of members here who will take a lot of time to explain a concept and won't stop until you're happy
    It makes perfect sense now thank you!. I sure will, it's my first time using the site and i got a reply after only 30mins. Again; thank you so much for your effort with me )
 
 
 
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